Question

At constant pressure P the volume of a gas increases from $V_1$ to $V_2$ when "Q amount of heat is removed from the system. What will happen to internal energy?

• A. Decrease by amount P$V_2$ - $V_1$-Q
• B. Remain the same if Q= P($V_1$-$V_2$)
• C. Increase by some unknown amount
• D. Decrease by an amount Q-P($V_1$-$V_2$)

Answer 1

Answer: (D)

Decrease by an amount Q-P($V_1$-$V_2$)

Answer 2

The problem involves the application of the First Law of Thermodynamics.

1. First Law of Thermodynamics:
   The First Law of Thermodynamics states that the change in internal energy ($\Delta U$) of a system is equal to the heat added to the system ($Q_{in}$) plus the work done on the system ($W_{on}$): $\Delta U = Q_{in} + W_{on}$

2. Analyze the given information and assign signs:

   • Heat (Q): "Q amount of heat is removed from the system."
     When heat is removed from the system, $Q_{in}$ is negative. So, $Q_{in} = -Q$.
   • Work (W): "volume of a gas increases from $V_1$ to $V_2$ at constant pressure P."
     For a process at constant pressure, the work done by the system ($W_{by}$) is given by: $W_{by} = P(V_2 - V_1)$
     Since the volume increases ($V_2 > V_1$), $V_2 - V_1$ is positive, meaning the system does positive work on the surroundings.
     The work done on the system ($W_{on}$) is the negative of the work done by the system: $W_{on} = -W_{by} = -P(V_2 - V_1)$

3. Substitute into the First Law:
   Substitute the expressions for $Q_{in}$ and $W_{on}$ into the First Law equation: $\Delta U = (-Q) + (-P(V_2 - V_1))$
   $\Delta U = -Q - P(V_2 - V_1)$

4. Interpret the result:

   • Since $Q$ is an amount of heat (a positive value), $-Q$ is a negative term.
   • Since $P$ is pressure (positive) and $(V_2 - V_1)$ is positive (volume increases), $-P(V_2 - V_1)$ is also a negative term.
   • Therefore, $\Delta U$ is the sum of two negative terms, which means $\Delta U$ is negative.
   • A negative $\Delta U$ indicates that the internal energy of the system decreases.

5. Calculate the amount of decrease:
   The amount by which the internal energy decreases is the magnitude of $\Delta U$: $|\Delta U| = |-Q - P(V_2 - V_1)| = |-(Q + P(V_2 - V_1))| = Q + P(V_2 - V_1)$

6. Compare with the options:
   Let's check Option 4: "Decrease by an amount Q-P($V_1$-$V_2$)"
   The amount of decrease stated in this option is $Q - P(V_1 - V_2)$.
   We know that $P(V_1 - V_2) = -P(V_2 - V_1)$.
   Substituting this into the option's expression:
   Amount of decrease $= Q - (-P(V_2 - V_1))$
   Amount of decrease $= Q + P(V_2 - V_1)$

This matches our derived amount of decrease.

Therefore, the internal energy decreases by an amount $Q - P(V_1 - V_2)$.